bài này dễ lắm,mình giải đây:

C = \(\frac{1}{100}\)- \(\frac{1}{100.99}\)-\(\frac{1}{99.98}\)\(\frac{1}{98.97}\)- ... - \(\frac{1}{3.2}\)- \(\frac{1}{2.1}\)

C = \(\frac{-1}{1.2}\)+ \(\frac{-1}{2.3}\) + ... +\(\frac{-1}{98.99}\)+ \(\frac{1}{99.100}\)+ \(\frac{1}{100}\)

C = \(\frac{-1}{1}\)- \(\frac{-1}{2}\)

Mình bận rồi , phần sau tự làm nha.

\(\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{2}-1\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-1\right)\)

\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}+1\)

\(=\frac{9799}{9900}\)

Gọi A=\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

A= -(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\))

A=-(1-\(\frac{1}{100}\))

A=-(\(\frac{99}{100}\))

A=-99/100

\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Leftrightarrow-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(\Leftrightarrow\)\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Leftrightarrow-\left(1-\frac{1}{100}\right)\)

\(\Leftrightarrow-\left(\frac{99}{100}\right)\)

\(=-\frac{99}{100}\)

=1/100+1/99-1/99+1/98-1/98+1/97-...........-1/2+1/2-1/2+1

=1/100+1

=101/100